malloc-ating multidimensional array in function

Question

I'm trying to allocate a 2d array in a C program. It works fine in the main function like this (as explained here):

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char ** argv)
{
    int ** grid;
    int i, nrows=10, ncols=10;
    grid = malloc( sizeof(int *) * nrows);

    if (grid == NULL){
        printf("ERROR: out of memory\n");
        return 1;
    }

    for (i=0;i<nrows;i++){
        grid[i] = malloc( sizeof(int) * ncols);
        if (grid[i] == NULL){
            printf("ERROR: out of memory\n");
            return 1;
        }
    }
    printf("Allocated!\n");

    grid[5][6] = 15;
    printf("%d\n", grid[5][6]);
    return 0;
}

But since I have to do this several times with different arrays, I was trying to move the code into a separate function.

#include <stdio.h>
#include <stdlib.h>

int malloc2d(int ** grid, int nrows, int ncols){
    int i;
    grid = malloc( sizeof(int *) * nrows);

    if (grid == NULL){
        printf("ERROR: out of memory\n");
        return 1;
    }

    for (i=0;i<nrows;i++){
        grid[i] = malloc( sizeof(int) * ncols);
        if (grid[i] == NULL){
            printf("ERROR: out of memory\n");
            return 1;
        }
    }
    printf("Allocated!\n");
    return 0;
}

int main(int argc, char ** argv)
{
    int ** grid;

    malloc2d(grid, 10, 10);
    grid[5][6] = 15;
    printf("%d\n", grid[5][6]);
    return 0;
}

However, although it doesn't complain while allocating, I get segmentation fault when accessing the array. I read different posts on decayed arrays and similar topics, but I still can't figure out how to solve this problem. I imagine I'm not passing the 2d array correctly to the function.

Many thanks.

Answer 1

That is not a multidimensional array; it is a single dimensional array containing pointers to single dimensional arrays. Multidimensional arrays do not contain pointers; they are single memory blocks.

Your problem here is that you have a pointer to a pointer, and you're trying to return it from your function through a parameter. If you're going to do that, you're going to need a pointer to a pointer to a pointer as your parameter, and you're going to have to pass the address of a pointer to a pointer to the method. If you don't do this, you're not changing the value of the variable grid in main -- you're changing the one which was copied as the parameter to the malloc2d function. Because the grid in main is left uninitialized, you get a undefined behavior.

Here's an example of what I mean as the fix:

#include <stdio.h>
#include <stdlib.h>

int malloc2d(int *** grid, int nrows, int ncols){
    int i;
    *grid = malloc( sizeof(int *) * nrows);

    if (*grid == NULL){
        printf("ERROR: out of memory\n");
        return 1;
    }

    for (i=0;i<nrows;i++){
        (*grid)[i] = malloc( sizeof(int) * ncols);
        if ((*grid)[i] == NULL){
            printf("ERROR: out of memory\n");
            return 1;
        }
    }
    printf("Allocated!\n");
    return 0;
}

int main(int argc, char ** argv)
{
    int ** grid;

    malloc2d(&grid, 10, 10);
    grid[5][6] = 15;
    printf("%d\n", grid[5][6]);
    return 0;
}

Additional notes:

• If a single allocation fails, you leak the allocation for the first array, as well as the allocations for all previous rows. You need to call free on those before returning.
• You're returning through a parameter, even though you really don't have to. If I were writing this, I'd make the method return int **, and signal error by returning 0.